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It’s now time to see our first major application of
derivatives in this chapter. Given a
continuous function, f(x), on an
interval [a,b] we want to determine
the absolute extrema of the function. To
do this we will need many of the ideas that we looked at in the previous
section.

First, since we have an interval and we are assuming that
the function is continuous the Extreme Value
Theorem tells us that we can in fact do this. This is a good thing of course. We don’t want to be trying to find something
that may not exist.

Next, we saw in the previous section that absolute extrema
can occur at endpoints or at relative extrema.
Also, from Fermat’s Theorem we know
that the list of critical points is also a list of all possible relative
extrema. So the endpoints along with the
list of all critical points will in fact be a list of all possible absolute
extrema.

Now we just need to recall that the absolute extrema are
nothing more than the largest and smallest values that a function will take so
all that we really need to do is get a list of possible absolute extrema, plug
these points into our function and then identify the largest and smallest
values.

Here is the procedure for finding absolute extrema.

Finding Absolute
Extrema of f(x) on [a,b].

Verify
that the function is continuous on the interval [a,b].

Find
all critical points of f(x)
that are in the interval [a,b]. This makes sense if you think about
it. Since we are only interested
in what the function is doing in this interval we don’t care about
critical points that fall outside the interval.

Evaluate
the function at the critical points found in step 1 and the end points.

Identify
the absolute extrema.

There really isn’t a whole lot to this procedure. We called the first step in the process step
0, mostly because all of the functions that we’re going to look at here are
going to be continuous, but it is something that we do need to be careful
with. This process will only work if we
have a function that is continuous on the given interval. The most labor intensive step of this process
is the second step (step 1) where we find the critical points. It is also important to note that all we want
are the critical points that are in the interval.

Let’s do some examples.

Example 1 Determine
the absolute extrema for the following function and interval.

Solution

All we really need to do here is follow the procedure
given above. So, first notice that
this is a polynomial and so in continuous everywhere and in particular is
then continuous on the given interval.

Now, we need to get the derivative so that we can find the
critical points of the function.

It looks like we’ll have two critical points, and . Note that we actually want something more
than just the critical points. We only
want the critical points of the function that lie in the interval in
question. Both of these do fall in the
interval as so we will use both of them.
That may seem like a silly thing to mention at this point, but it is
often forgotten, usually when it becomes important, and so we will mention it
at every opportunity to make sure it’s not forgotten.

Now we evaluate the function at the critical points and
the end points of the interval.

Absolute extrema are the largest and smallest the function
will ever be and these four points represent the only places in the interval
where the absolute extrema can occur.
So, from this list we see that the absolute maximum of g(t) is 24 and it occurs at (a critical point) and the absolute minimum
of g(t) is -28 which occurs at (an endpoint).

In this example we saw that absolute extrema can and will
occur at both endpoints and critical points.
One of the biggest mistakes that students make with these problems is to
forget to check the endpoints of the interval.

Example 2 Determine
the absolute extrema for the following function and interval.

Solution

Note that this problem is almost identical to the first
problem. The only difference is the
interval that we’re working on. This
small change will completely change our answer however. With this change we have excluded both of
the answers from the first example.

The first step is to again find the critical points. From the first example we know these are and .. At this point it’s important to recall that
we only want the critical points that actually fall in the interval in
question. This means that we only want
since falls outside the interval.

Now evaluate the function at the single critical point in
the interval and the two endpoints.

From this list of values we see that the absolute maximum
is 8 and will occur at and the absolute minimum is -3 which occurs
at .

As we saw in this example a simple change in the interval
can completely change the answer. It
also has shown us that we do need to be careful to exclude critical points that
aren’t in the interval. Had we forgotten
this and included we would have gotten the wrong absolute
maximum!

This is the other big mistakes that students make in these
problems. All too often they forget to
exclude critical points that aren’t in the interval. If your instructor is anything like me this
will mean that you will get the wrong answer. It’s not too hard to make sure that a critical
point outside of the interval is larger or smaller than any of the points in
the interval.

Example 3 Suppose
that the population (in thousands) of a certain kind of insect after t months is given by the following
formula.

Determine the minimum and maximum population in the first
4 months.

Solution

The question that we’re really asking is to find the
absolute extrema of P(t) on the
interval [0,4]. Since this function is
continuous everywhere we know we can do this.

Let’s start with the derivative.

We need the critical points of the function. The derivative exists everywhere so there
are no critical points from that. So,
all we need to do is determine where the derivative is zero.

The solutions to this are,

So, these are all the critical points. We need to determine the ones that fall in
the interval [0,4]. There’s nothing to
do except plug some n’s into the
formulas until we get all of them.

:

We’ll need both of these critical points.

:

We’ll need these.

:

In this case we only need the first one since the second
is out of the interval.

There are five critical points that are in the
interval. They are,

Finally, to determine the absolute minimum and maximum
population we only need to plug these values into the function as well as the
two end points. Here are the function
evaluations.

From these evaluations it appears that the minimum
population is 100,000 (remember that P
is in thousands…) which occurs at and the maximum population is 111,900 which
occurs at .

Make sure that you can correctly solve trig equations. If we had forgotten the we would have missed the last three critical
points in the interval and hence gotten the wrong answer since the maximum
population was at the final critical point.

Also, note that we do really need to be very careful with
rounding answers here. If we’d rounded
to the nearest integer, for instance, it would appear that the maximum
population would have occurred at two different locations instead of only one.

Example 4 Suppose
that the amount of money in a bank account after t years is given by,

Determine the minimum and maximum amount of money in the
account during the first 10 years that it is open.

Solution

Here we are really asking for the absolute extrema of A(t) on the interval [0,10]. As with the previous examples this function
is continuous everywhere and so we know that this can be done.

We’ll first need the derivative so we can find the
critical points.

The derivative exists everywhere and the exponential is
never zero. Therefore the derivative
will only be zero where,

We’ve got two critical points, however only is actually in the interval so that is only
critical point that we’ll use.

Let’s now evaluate the function at the lone critical point
and the end points of the interval.
Here are those function evaluations.

So, the maximum amount in the account will be $2000 which
occurs at and the minimum amount in the account will
be $199.66 which occurs at the 2 year mark.

In this example there are two important things to note. First, if we had included the second critical
point we would have gotten an incorrect answer for the maximum amount so it’s
important to be careful with which critical points to include and which to
exclude.

All of the problems that we’ve worked to this point had
derivatives that existed everywhere and so the only critical points that we
looked at were those for which the derivative is zero. Do not get too locked into this always
happening. Most of the problems that we
run into will be like this, but they won’t all be like this.

Let’s work another example to make this point.

Example 5 Determine
the absolute extrema for the following function and interval.

Solution

Again, as with all the other examples here, this function
is continuous on the given interval and so we know that this can be done.

First we’ll need the derivative and make sure you can do
the simplification that we did here to make the work for finding the critical
points easier.

So, it looks like we’ve got two critical points.

Both of these are in the interval so let’s evaluate the
function at these points and the end points of the interval.

The function has an absolute maximum of zero at and the function will have an absolute
minimum of -15 at .

So, if we had ignored or forgotten about the critical
point where the derivative doesn’t exist ( ) we would not have gotten the
correct answer.

In this section we’ve seen how we can use a derivative to
identify the absolute extrema of a function.
This is an important application of derivatives that will arise from
time to time so don’t forget about it.